The masses m each are placed at the corners of an equilateral triangle of side a and a mass $\frac{m}{3}$ is placed at the centre of the centre of the triangle. The 3 masses are in uniform circular motion with velocity $v$ on the circumcircle of the triangle, the centripetal force being provided by their mutual attraction. Then

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$v = \sqrt{\frac{GM}{a} \frac{(\sqrt{3} + 1)}{\sqrt{3}}}$

$v = \sqrt{\frac{GM}{a} \frac{(\sqrt{3} - 1)}{\sqrt{3}}}$

The mass $\frac{m}{3}$ experiences no force

If $\frac{m}{3}$ is absent, $v = \sqrt{\frac{Gm}{a}}$

answer is A, C, D.

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Detailed Solution

The resultant of all forces on each m acts in
the direction towards centre.

$F_{r} = F^{'} + 2 F \cos 30^{\circ} = F^{'} + \sqrt{3} F$

$= \frac{Gm \frac{M}{3}}{a^{2} / 3} + \frac{\sqrt{3} Gmm}{a^{2}} = \frac{{Gm}^{2}}{a^{2}} [1 + \sqrt{3}]$

Since this provides the centripetal force:

$\frac{{Gm}^{2}}{a^{2}} (1 + \sqrt{3}) = \frac{{mv}^{2}}{(\frac{a}{\sqrt{3}})}$

$∴ v = \sqrt{\frac{Gm}{a} (\frac{1 + \sqrt{3}}{\sqrt{3}})} \Rightarrow If \frac{m}{3}$ is absent, then

$F_{r} = \sqrt{3} F \Rightarrow∴ \sqrt{3} \frac{Gmm}{a^{2}} = \frac{{mv}^{2}}{\frac{a}{\sqrt{3}}} \Rightarrow v = \sqrt{\frac{Gm}{a}}$

Resultant force on $\frac{m}{3} \Rightarrow 0$

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